Question

1. pat and chris both took a spatial abilities test (mean = 80, std. dev. = 8). pat scored a 76 and chris scored a 94. what proportion of individuals would score between pat and chris?

Explanation:

Step1: Calculate z - score for Pat

The formula for the z - score is $z=\frac{X - \mu}{\sigma}$, where $X$ is the raw score, $\mu$ is the mean, and $\sigma$ is the standard deviation. For Pat, $X = 76$, $\mu=80$, $\sigma = 8$.
So, $z_{Pat}=\frac{76 - 80}{8}=\frac{- 4}{8}=- 0.5$

Step2: Calculate z - score for Chris

For Chris, $X = 94$, $\mu = 80$, $\sigma=8$.
Using the z - score formula, $z_{Chris}=\frac{94 - 80}{8}=\frac{14}{8}=1.75$

Step3: Find the proportion corresponding to z - scores

We use the standard normal distribution table (z - table).

• The proportion of values less than $z=- 0.5$ is $P(Z < - 0.5)=0.3085$ (from z - table).
• The proportion of values less than $z = 1.75$ is $P(Z < 1.75)=0.9599$ (from z - table).

To find the proportion of values between Pat and Chris, we subtract the two proportions: $P(-0.5$P(-0.5

Answer:

The proportion of individuals who would score between Pat and Chris is $0.6514$ (or $65.14\%$)